# Thread: Help with basic geometery problem

1. ## Help with basic geometery problem

Not sure what I'm doing wrong...

image1.jpg

2. Originally Posted by Sparklez
Not sure what I'm doing wrong...

image1.jpg
Cannot read the attachment - not sure what you are doing!!

3. I see (after twisting my neck!) "right solid".
What does that mean?

4. Originally Posted by Denis
I see (after twisting my neck!) "right solid".
What does that mean?
That generally means the polar axis of the cross-section is vertical - like right-circular cylinder, right-circular cone, right-rectangular prism, etc.

5. Originally Posted by Sparklez
Not sure what I'm doing wrong...
Why do you think that you're doing something wrong?

Originally Posted by Sparklez
Your image is sideways (hard to read) and waaaaaaay too small. I think the written portion of the image says the following:

d) right solid

Area of bases --> 24 x 2 = 48 8+2
. . . . . . . . .= 2pi[something]2 (2) = 12.56

A = pi (26+)2 = 2x(1+2)
6 x 4 = 245+4

Area of book --> 2062 22 = 40102
S24ft2 = 20ft2

Area of cylinder --> 91.4ft2

half [something] circum. (2pi[something])/(3a2) = 6.28ft x 5 = 31.42t

SKS = 6 x 5(2) = 60ft2

91.4 + 40 + 12.6 + 48ft2 = 191.96ft2

Naturally, we're having a bit of difficulty with this.

You don't include the original exercise (such as the instructions); would the following be a correct statement of the question?

The shape shown below is the top view of a right solid:

Code:
top view:

+---------+..
|         |  .
|4'       |   |
|         |  ,'
+---------+''
6'`
The solid has a height of five feet. (Note: All units are in "feet".) The rounded portion at the right of the above shape is a semicircle with radius two feet.

Given this information, find the surface area of the solid.

There are two rectangular areas (one on the top and one on the bottom) of this prism. Use the given length (6') and depth (4') to find the area of one of these rectangles, and then multiply by 2. Keep track of this by labelling this portion as, say, "top & bottom rectangles".

There is one semicircular area on the diagram of the top of the prism; there is a corresponding one on the bottom. Together, these make one circle with the given radius (2'). Plug this into the formula for the area of a circle. (Do NOT round the result! Keep it in terms of "pi"!) Keep track of this value by labelling it as, say, "circular area".

On the left-hand side of the drawing is a line which represents a vertical "side" of the wrap-around portion of this solid (being the part that is not the bottom of the solid and is the part that we cannot see in the drawing). You have a depth (4') and a height (5') for this rectangular area. Plug these dimensions into the formula for the area of a rectangle. Keep track of this value by labelling it, say, "left-hand side".

On the "upper edge" in the drawing, you have another vertical "side". What is the length of this? What is the height of this? What value does this then give you for the area of that rectangular side? How does this relate to the area of the rectangular side represented by the "lower edge" in the drawing? Keep track of this by labelling it, say, "upper & lower rectangles".

The remainder of the unseen wrap-around is half of the side-surface-area of a cylinder with radius 2' and height 5' (that is, it is the part of the surface area which does not include the end-caps). Plug these values into the formula for the side-surface-area of a right circular cylinder, and divide by two to get the area you need. Label this as, say, "cylindrical part".

Add the values, keeping everything exact (that is, in terms of "pi", not decimal round-offs). What do you get?

If you get stuck, please reply with a clear listing of your steps and results in following the above step-by-step instructions. Thank you!

6. Thank you so much! But I'm not sure why I shouldn't round off pi? How to I figure out the surface area without doing that? I tried to redo the problem following your steps best I could, and I came up with I different answer. Every time. That's why I don't understand what mistake I'm making, and if it's the rounding off pi could you help me understand why that is? Thanks! Here is the problem redone:
image1.jpg

7. Originally Posted by Sparklez
Thank you so much!
Does this mean that I correctly guessed the exercise? If not, then my step-by-step instructions are wrong. Please confirm or correct.

Originally Posted by Sparklez
But I'm not sure why I shouldn't round off pi?
To avoid decimal round-offs, and the resulting round-off errors.

Originally Posted by Sparklez
How to I figure out the surface area without doing that?
Do the computations in the usual way. Leave "$\pi$" as "$\pi$". It's just like if you were multiplying 3 and 5, but they told you to "write the product, but do not simplify", so you'd end up with 3*5 rather than 15; or if you're adding terms like 2x and 3, so you end up with "2x + 3", because the terms are unlike and cannot be combined.

Originally Posted by Sparklez
I tried to redo the problem following your steps best I could, and I came up with I different answer. Every time. That's why I don't understand what mistake I'm making, and if it's the rounding off pi could you help me understand why that is? Thanks! Here is the problem redone:
image1.jpg
Unfortunately, this is a very small image, so I can't be sure what you're doing. (For instance, it looks as though you counted the semicircular regions from the top and bottom twice, rather than following the instructions provided.) Also, since we don't know what the target answer is, we cannot comment on whether round-off errors may be contributing to your issues.

When you reply, please type out at least one example of your steps and answers, and list at least two or three of the various different answers you got. Thank you!

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